SATHEE: Equilibrium - Result Question 82

Equilibrium - Result Question 82

####86. What is $[H^{+}]$ in $m o l / L$ of a solution that is $0.20 M$ in $C H_{3} C O O N a$ and $0.10 M$ in $C H_{3} C O O H$ ? $K_{a}$ for $C H_{3} C O O H = 1.8 \times 10^{- 5}$ .

[2010]

(a) $3.5 \times 10^{- 4}$

(b) $1.1 \times 10^{- 5}$

(d) $9.0 \times 10^{- 6}$

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Solution:

(d) $p H = p K_{a} + \log [\frac{Salt}{Acid}]$

$\log [H^{+}] = \log K_{a} - \log [\frac{Salt}{Acid}]$

$\log [H^{+}] = \log K_{a} + \log [\frac{Acid}{Salt}]$

$\begin{aligned} [H^{+}] & = K_{a} [\frac{Acid}{Salt}] \\ = 1.8 \times 10^{- 5} \times \frac{0.1}{0.2} = 9 \times 10^{- 6} M \end{aligned}$

Equilibrium - Result Question 82

Solution:

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Equilibrium - Result Question 82

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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